# Why is the damping force on a spring oscillator linearly dependent on velocity?

If you consider the damping force is friction like in:

then the force should be $$F=\mu N$$ where $\mu$ is the coefficient of kinetic friction. Why then is the damping force assumed to be linearly dependent on velocity?

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The damping on a linear spring-mass system is usually assumed to be due to a dashpot. Dashpots force a fluid through a small orifice and the resulting force is proportional to the velocity. –  OSE Apr 26 '13 at 22:38
Cars these days have shock absorbers that are basically dashpots with viscous drag. There was a time when cars used straight friction devices to absorb the energy in up-and-down wheel motions. It just depends what type of energy-absorbing device you use. –  Mike Dunlavey Apr 26 '13 at 23:38
You are right that Coulomb damping is what you need to describe sliding motion. A more accurate model in the case of sliding friction includes both static and kinetic friction: stick-slip friction, which in some cases can approximate a velocity dependent law, but the resulting differential equation must be solved numerically. –  Michael Brown Apr 26 '13 at 23:39

This question is actually one of the lab exercises I teach. For a spring-mass system, if the damping force is friction, then it is independent of velocity (verified experimentally). However, as mentioned in the comments, the damping force may not always be friction. For example, if the mass is a material like aluminium and it is oscillating over some magnets, the damping force will be linearly dependent on velocity.

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I wrote a program to approximate an oscillator of this sort (with friction as the damping force) and it surprised me that the mass did NOT return to its equilibrium position! Instead it came to rest at a point offset from equilibrium by a small amount. –  Greg Apr 26 '13 at 22:51
That is what can happen with velocity independent damping. The opposing force is constant as long as it's moving, so it stops wherever it is when the acceleration makes it stop. If damping is dependent on velocity, then it alsways returns to equilibrium position. –  Jimdalf the Grey Apr 26 '13 at 22:56
@ Greg : yes that's because that kind of friction has a 'threshold'... one can see it from the phase diagram plot where you will see spirals towards the center but it will stop somewhere on the x-axis not on the origin in general. –  nervxxx Apr 26 '13 at 22:56

Actually, this is just a simple model for resistive forces (usually duo to viscosity) , and in many situations you can not assume this force to be linearly dependent on velocity.(for example ,usually this model is correct only for small enough objects)

In many everyday examples, this force is due to viscous forces. If you consider an (small enough) object moving in a viscose fluid ,and if the speed of the object is small enough, the fluid particles will move parallel to it, and their speed will vary linearly from $v$ in the vicinity of the object to zero in farther points. Each layer of fluid will move faster than the one just outer , and friction between them will give rise to a force resisting the motion of object. (a force in the direction opposite to its motion.)

There is a formula (named Stokes' law) for the frictional force on such a small (spherical) particle moving with constant(terminal) speed in a viscose fluid :

$F = 6 \pi\mu Rv$

where $v$ is the particle's velocity and $R$ is the radius of the spherical object and $\mu$ is a measure of fluid's viscosity.(named dynamic viscosity)

A linear model for resistive forces may be used in other contexts too; One can model laser cooling process as a linear resistive force against the motion of microscopic particles.

For more quantitative formulas about viscous forces you can see Viscosity in Wikipedia.

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Refer to Stokes's law too :) –  nervxxx Apr 26 '13 at 22:59